I am quite stuck here as to how my book achieves an answer.
The question reads that there are two boxes, lets call them $A$ and $B.$ Box $A$ has $24$ balls; $3$ red, $8$ yellow, $13$ green. Box $B$ has $18$ balls; $5$ red, $7$ yellow, $6$ green.
the question ask: what is the probability of choosing one ball from each box, such that both balls are the same color$?$
Once again I understand that P(event)= outcomes of event/ total outcomes
now the book states the answer as being $149/432$, but I do not know how they did so.
I know that my sample space is $432$, because there is an intersection from box $A$ and $B$; such that $24\times 18$ yields the total possible outcomes. So I got $432$ part.
Then if I pick a red from box $A$ then I have a $5/18$ chance picking red from $B$ if I picked a yellow from box $A$ then I have a $7/18$ chance of picking yellow from $B$ and if I picked a green from box $A$ then I have a $6/18$ chance of picking a green. But I don't know how to use this information to my advantage in order to solve it